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求解这个方程组嗯。

查看数: 2313 | 评论数: 6 | 收藏 0
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    组图打开中,请稍候......
发布时间: 2012-8-14 09:16

正文摘要:

本帖最后由 格物知理 于 2012-8-23 09:16 编辑 呢?求解,谢谢。

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格物知理 发表于 2012-8-27 17:05:14
。。。。。。
格物知理 发表于 2012-8-23 08:56:06
kuafuzhuiri 发表于 2012-8-22 20:48
真复杂。。

是很复杂{:F16:}
kuafuzhuiri 发表于 2012-8-22 20:48:20
真复杂。。
格物知理 发表于 2012-8-15 18:16:01
木有人吗。。。
格物知理 发表于 2012-8-14 16:35:30
杨梦如 发表于 2012-8-14 11:41
% MathType!MTEF!2!1!+-
% feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBae ...

这是。。?{:F36:}
杨梦如 发表于 2012-8-14 11:41:05
% MathType!MTEF!2!1!+-
% feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqiVu0Je9sqqr
% pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs
% 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai
% aabeqaamaabaabauaakqaabeqaaiaamgdacaaJTaGaaWinaiaam6ca
% caaJ+UOaamyyaiaacYcacaWGIbGaaiilaiaadogacqGH+aGpcaaIWa
% GaaiilaiaamkeBcaaJbVOaaiOoaaqaaiaacIcacaaIXaGaey4kaSYa
% aSaaaeaacaWGHbaabaGaamOyaaaacaGGPaGaaiikaiaaigdacqGHRa
% WkdaWcaaqaaiaadkgaaeaacaWGJbaaaiaacMcacaGGOaGaaGymaiab
% gUcaRmaalaaabaGaam4yaaqaaiaadggaaaGaaiykaiabgwMiZkaaik
% dacaGGOaGaaGymaiabgUcaRmaalaaabaGaamyyaiabgUcaRiaadkga
% cqGHRaWkcaWGJbaabaWaaOqaaeaacaWGHbGaamOyaiaadogaaSqaai
% aaiodaaaaaaOGaaiykaiaac6caaeaadaWcaaqaaiaacIcacaaIXaGa
% ey4kaSYaaSaaaeaacaWGHbaabaGaamOyaaaacaGGPaGaaiikaiaaig
% dacqGHRaWkdaWcaaqaaiaadkgaaeaacaWGJbaaaiaacMcacaGGOaGa
% aGymaiabgUcaRmaalaaabaGaam4yaaqaaiaadggaaaGaaiykaiabgk
% HiTiaaikdaaeaacaaIYaaaaiabg2da9maalaaabaGaamyyaaqaaiaa
% dkgaaaGaey4kaSYaaSaaaeaacaWGIbaabaGaam4yaaaacqGHRaWkda
% WcaaqaaiaadogaaeaacaWGHbaaaiabgwMiZoaalaaabaGaamyyaiab
% gUcaRiaadkgacqGHRaWkcaWGJbaabaWaaOqaaeaacaWGHbGaamOyai
% aadogaaSqaaiaaiodaaaaaaaaaaa!8E49!
\[\begin{array}{l}
1 - 4.a,b,c > 0,:\\
(1 + \frac{a}{b})(1 + \frac{b}{c})(1 + \frac{c}{a}) \ge 2(1 + \frac{{a + b + c}}{{\sqrt[3]{{abc}}}}).\\
\frac{{(1 + \frac{a}{b})(1 + \frac{b}{c})(1 + \frac{c}{a}) - 2}}{2} = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge \frac{{a + b + c}}{{\sqrt[3]{{abc}}}}
\end{array}\]

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